When submitting a solution in C++, please select either C++14 (GCC 6-32) or C++17 (GCC 7-32) as your compiler. ×

kittyK's blog

By kittyK, history, 4 years ago, In English

Can anyone explain how to solve this 687B - Remainders Game . I don't understand how to come with ideas for such problems and I did not understand the editorial well.

  • Vote: I like it
  • +2
  • Vote: I do not like it

»
4 years ago, # |
  Vote: I like it 0 Vote: I do not like it

If you know xmodc1,xmodc2,...,xmodcn then you can find out xmodlcm(c1,c2,...,cn) if the given values are not self-contradictory(eg. xmod2=0 and xmod4=1). Conversely, if you know xmodlcm(c1,c2,...,cn) you can find each of the given n values. So first find xmodlcm(c1,c2,...,cn) and this would retain all the information. Now, if k divides lcm(x1,x2,...,xn) you can easily find xmodk otherwise you can't.

»
4 years ago, # |
  Vote: I like it +10 Vote: I do not like it

The editorial proves that all you have to do is check whether $$$k \mid \text{lcm}(c_1, \dots, c_n)$$$. This is equivalent to saying $$$\gcd(k, \text{lcm}(c_1, \dots, c_n)) = k$$$. But we can calculate the left side very easily since $$$\gcd(k, \text{lcm}(c_1, \dots, c_n)) = \text{lcm}(\gcd(k, c_1), \dots, \gcd(k, c_n))$$$. The time complexity will be $$$O(n(\log(C) + \log(K)))$$$. Also there will not be overflow because the intermediate value will always be $$$\leq k$$$.

These identities hold because if you think of an integer's prime factorization as a multiset, $$$\mid$$$ corresponds to $$$\subseteq$$$, $$$\gcd$$$ corresponds to $$$\cap$$$, and $$$\text{lcm}$$$ corresponds to $$$\cup$$$